1. Prove by example the following relational algebra property: (5 points) ((R))=(R) as long as contains the attributes in You may use the COMPANY schema or another other schema of your choice. Figure 5.6 One possible database state for the COMPANY relational database schema. EMPLOYEE \begin{tabular}{|l|c|l|c|c|c|c|c|c|c|} \hline Frame & Minit & Lname & Sen & Bdate & Address & Sex & Saley & Super_ssn & Dno \\ \hline John & B & Smith & 123456789 & 19650109 & 731 Fandren, Houston, TX & M & 30000 & 333445555 & 5 \\ \hline Franklin & T & Wong & 333446565 & 19651208 & 638 Vosa, Houaton, TX & M & 40000 & 888665565 & 5 \\ \hline Alicie & J & Zelaya & 999887777 & 19680119 & 3321 Cestle, Spring. TX & F & 25000 & 987654321 & 4 \\ \hline Jenniles & S & Wallace & 987654321 & 19410620 & 281 Berry, Bellaire, TX & F & 43000 & 888665555 & 4 \\ \hline Rameeh & K & Narayan & 666884444 & 19620915 & 975 Fire Oak, Humble, TX & M & 38000 & 333445555 & 5 \\ \hline Joyce & A & English & 453453453 & 19720731 & 5631 Rice, Houston, TX & F & 25000 & 333445555 & 5 \\ \hline Ahmad & V & Jabbar & 987987987 & 19690329 & 980 Dallas, Houston, TX & M & 25000 & 987654321 & 4 \\ \hline James & E & Borg & 888865555 & 19371110 & 450 Stone, Houston, TX & M & 55000 & NULL & 1 \\ \hline \end{tabular} DEPARTMENT DEPT_LOCATIONS \begin{tabular}{|l|c|c|c|} \hline \multicolumn{1}{|c|}{ Dname } & Drumber & Mgr_ssn & Mgr_start_date \\ \hline Research & 5 & 333445555 & 19880522 \\ \hline Administration & 4 & 987654321 & 19950101 \\ \hline Headquartors & 1 & 888665585 & 198106.19 \\ \hline \end{tabular} \begin{tabular}{|c|l|} \hline Dnumber & Dlocation \\ \hline 1 & Houston \\ \hline 4 & Stafford \\ \hline 5 & Bollaire \\ \hline 5 & Sugarlend \\ \hline 5 & Houston \\ \hline \end{tabular} WORKS ON PROJECT \begin{tabular}{|c|c|c|} \hline Essn & Pno & Houre \\ \hline 123456789 & 1 & 32.5 \\ \hline 123456789 & 2 & 7.5 \\ \hline 686884444 & 3 & 40.0 \\ \hline 453453453 & 1 & 20.0 \\ \hline 483483483 & 2 & 20.0 \\ \hline 333446555 & 2 & 10.0 \\ \hline 393445555 & 3 & 10.0 \\ \hline 333445555 & 10 & 10.0 \\ \hline 333445555 & 20 & 10.0 \\ \hline 909837777 & 30 & 30.0 \\ \hline 998887777 & 10 & 10.0 \\ \hline 987987987 & 10 & 35.0 \\ \hline 987987987 & 30 & 5.0 \\ \hline 987654321 & 30 & 20.0 \\ \hline 987654321 & 20 & 15.0 \\ \hline 888665555 & 20 & NULL \\ \hline \end{tabular} \begin{tabular}{|l|c|l|c|} \hline \multicolumn{1}{|c|}{ Prame } & Pnumber & Plocation & Dnum \\ \hline ProductX & 1 & Bellaire & 5 \\ \hline ProductY & 2 & Sugarland & 5 \\ \hline ProductZ & 3 & Houston & 5 \\ \hline Cornputerization & 10 & Stafford & 4 \\ \hline Reorganization & 20 & Houston & 1 \\ \hline Newbenetitn & 30 & Stafford & 4 \\ \hline \end{tabular} DEPENDENT \begin{tabular}{|l|l|c|c|l|} \hline Eesn & Dependent name & Sex & Bdate & Relationship \\ \hline 333445555 & Alice & F & 19860405 & Daughter \\ \hline 333445585 & Theodore & M & 19831025 & Son \\ \hline 333448585 & by & F & 19880503 & Spouse \\ \hline 987654321 & Abner & M & 19420228 & Spouse \\ \hline 123456789 & Mchael & M & 19880104 & Son \\ \hline 123456789 & Alice & F & 19881230 & Daughter \\ \hline 123456789 & Elababeth & F & 19670506 & Spouse \\ \hline \end{tabular}